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Does a single boost change the spin orientation?

  1. Jun 25, 2013 #1
    Say I have a particle in the rest frame with spin up in the z direction. If I boost the system in x direction, does the spin still point in the z direction? Why or why not?

    In Dirac Spinor, if I choose "spin up", namely setting [itex]\phi=\begin{bmatrix}1\\0\end{bmatrix}[/itex] in the notation shown in Wikipedia, does it mean that it's pointing in the positive z direction regardless of the momentum direction of the particle? Or in other words, what direction does a general Dirac spinor with "spin up" points to?
     
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  3. Jun 25, 2013 #2

    Avodyne

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    You have to make a Lorentz boost on the spinor. How to do this is explained in the David Miller notes referenced by the wikipedia page you cited (as well as many other texts). In general this will change the spin direction if it is not parallel to the momentum.
     
  4. Jun 25, 2013 #3
    Since the Dirac spinor can be thought as a spinor directly boosted from the rest frame, if your claim is true, does it mean that when we say "a spin-up Dirac spinor", we don't mean a spin that's pointing in the +z direction if the momentum is not parallel to z direction? Which direction is it then?

    I know how to do the boost, but how do I find out which direction is the spin pointing in after the boost? The spin operator in [itex]\hat{n}[/itex] direction: [itex]\vec{\Sigma}\cdot\hat{n}=\frac{1}{2}\begin{bmatrix}\vec{\sigma}\cdot \hat{n} & 0 \\ 0 & \vec{\sigma}\cdot \hat{n}\end{bmatrix}[/itex] doesn't work for boosted spinors, since no matter which direction [itex]\hat{n}[/itex] you choose, the boosted spinor will not be an eigenstate. One can see this by realizing that the first two components and the second two components of the boosted spinor can't be eigenstates of the same [itex]\vec{\sigma}\cdot \hat{n}[/itex].

    So what is the appropriate spin operator for boosted spinors? How can I see that the spin direction is changed?
     
  5. Jun 26, 2013 #4

    Avodyne

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    Your spin operator is the correct one. After boosting, in general you get a superposition of states of definite spin along any given axis. You only retain a spin eigenspinor if you boost in the direction of the original spin axis.
     
  6. Jun 26, 2013 #5

    Bill_K

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    The idea that the spin "points in a particular direction" is not valid. By taking a superposition of states, you can quantize the spin along any axis you choose. As Avodyne says, the easiest axis to quantize along in this case is along the direction of motion. The spin states are then the helicity states.
     
  7. Jun 26, 2013 #6

    DrDu

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    I would be careful with what is meant with spin vs. spinors.
    The direction of spin is the direction of the Pauli-Lyubarsky vector and not of the spinor whose direction does not change under a single boost (but eventually under a combination of several boosts along different directions).
     
  8. Jun 26, 2013 #7
    It seems to me different people are saying different things. Let me explain my confusion a little bit more.

    I understand that I can retain the spinor direction if I boost it in the direction of the spin from the rest frame. What I don't understand is how to find out the direction of a spinor that's boosted in some other direction from rest frame. And as I said before, there is no way you can write the boosted spinor as any mixture of the eigenstates of that spin operator, so I don't think that spin operator is the correct one. Another possibility is proposed by Bill_K:

    But I can't make sense of a spin that doesn't point in a particular direction, especially since I started with a spin that points in a particular direction. I merely boosted the system to another frame, why should the spin now become ambiguous? I can reduce my original question to this: if I have a gyroscope pointing in +z direction in the rest frame, what direction is it pointing in if I see it from another inertia frame that's moving in say -x direction with velocity v? I would think this is the same question for our spin, since the lorentz transformation should work exactly the same on a group theory level.

    Since the direction of the helicity eigenstates are unambiguous, I attempted the following way to find out the direction of the boosted spinor. Please take a look and see whether this logic is right:

    First, Helicity spinors moving with momentum [itex]\mathbf{P}[/itex] can be obtained by first boosting a spin-up spinor in the +z direction (parallel to spin orientation) and then rotate to [itex]\mathbf{P}[/itex] direction (see Page 195 in Sakurai's Advanced Quantum Mechanics). If the particle is moving in the xz plane, the helicity spinor can be written as the following in the standard representation:

    [itex]
    \begin{align}
    u^{+}_{H}(\mathbf{P})=\sqrt{E+M}
    \begin{bmatrix}
    \cos\dfrac{\theta}{2}\\
    \sin\dfrac{\theta}{2}\\
    \dfrac{|\mathbf{P}|}{E+M}\cos\dfrac{\theta}{2}\\[6pt]
    \dfrac{|\mathbf{P}|}{E+M}\sin\dfrac{\theta}{2}
    \end{bmatrix},
    \quad
    u^{-}_{H}(\mathbf{P})=\sqrt{E+M}
    \begin{bmatrix}
    -\sin\dfrac{\theta}{2}\\
    \cos\dfrac{\theta}{2}\\
    \dfrac{|\mathbf{P}|}{E+M}\sin\dfrac{\theta}{2}\\[6pt]
    -\dfrac{|\mathbf{P}|}{E+M}\cos\dfrac{\theta}{2}
    \end{bmatrix},
    \end{align}
    [/itex]
    where [itex]\theta[/itex] is the angle between [itex]\mathbf{P}[/itex] and +z direction.

    Now we can easily write the Dirac spinor in standard representation as a combination of these two Helicity spinors:
    \begin{align}
    u^{+}_{D}(\mathbf{P})=\sqrt{E+M}
    \begin{bmatrix}
    1\\[8pt]
    0\\[8pt]
    \dfrac{P_{z}}{E+M}\\[8pt]
    \dfrac{P_{x}}{E+M}
    \end{bmatrix}
    =\sqrt{E+M}
    \begin{bmatrix}
    1\\[8pt]
    0\\[8pt]
    \dfrac{|\mathbf{P}|\cos\theta}{E+M}\\[8pt]
    \dfrac{|\mathbf{P}|\sin\theta}{E+M}
    \end{bmatrix}
    =\cos\dfrac{\theta}{2}u^{+}_{H}(\mathbf{P})-\sin\dfrac{\theta}{2}u^{-}_{H}(\mathbf{P}).
    \end{align}

    From this I can conclude that the above "spin-up" Dirac spinor is the "+" helicity spinor rotated by "-θ". And since the "+" helicity spinor is rotated from +z direction by "θ" angle, this Dirac spinor is again pointing in +z direction. So the above derivation demonstrates that Dirac spinor with "spin-up" always points in the +z direction. In other words, a "spin-up" spinor in the rest frame, when boosted to some arbitrary direction, remains pointing in the +z direction. One can prove similar results for spin down. This suggests that a single boost from rest frame doesn't change the spin orientation. And the gyroscope in a moving frame would still point in the same direction as in the rest frame.

    In order for the above logic be consistent, the spin operator for a moving particle has to be modified to [itex]e^{-i\boldsymbol\eta\mathbf{K}}\Sigma e^{i\boldsymbol\eta\mathbf{K}}[/itex], if the spinor is boosted by [itex]e^{-i\boldsymbol\eta\mathbf{K}}[/itex], namely the operator has to be transformed in the opposite way in order to maintain the same spin eigenvalue. It's as if this is just a matter of change of basis. Speaking of which, this logic seems to suggest that for every different momentum in different direction, we need to change the basis for the spin operator in order to get a sensible answer.

    Does this make sense? If my logic is correct, is there a simpler way to see why the spin orientation doesn't change direction under a single boost from the rest frame?

    I haven't thought this through yet, but one "contradiction" to the above conclusion can be presented as follows: the commutation relation [itex][J_{i}, K_{j}]=i\epsilon_{ijk}K_{k}[/itex] says that boost and rotation don't commute unless [itex]\mathbf{J}[/itex] and [itex]\mathbf{K}[/itex] are collinear, which suggest that if I first rotate a spin that's originally pointing in +z by θ, and then boost it in [itex]\mathbf{P}[/itex] direction, it would end up in a different state if I were to first boost it in P direction and then rotate the spinor by the same angle. Since the first set of operations gives me the helicity spinor, and the second one should differ, so when I first boost the spinor, it has to rotate by some amount so that later when I rotate by θ it doesn't end up in the same state. Is this right? I feel like something's wrong with this argument, but I can't put my finger on it yet. How does one resolve this contradiction?

    And last but not least is DrDu's response:

    The claim that the spinor direction does not change seems to agree with my above derivation. But I don't know what you mean by the direction of spin vs spinors. What does it mean that spin has a different direction than spinor? What do they separately represent? More explicitly, if I say I have a "spin-up" Dirac spinor, what do you say is the spin direction and spinor direction according to your definition? Which one is the physically observed one? For example, if I put a stream of high velocity electron through a Stern-Gerlach apparatus (the momentum P is approximately in horizontal direction but spins are detected in vertical direction), do the ones deflected up end up in a state that has spin up or spinor up?
     
    Last edited: Jun 26, 2013
  9. Jun 26, 2013 #8

    Bill_K

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    :eek: You need to review your understanding of angular momentum!

    An electron has only two spin states. If you quantize spin along the z-axis, the spin-up and spin-down states do not point along the z-axis! Comparing a spinning electron to a classical gyroscope is totally invalid. If you take the spin-up state and ask for its projection along some other axis, it will be a superposition of those two states.
     
  10. Jun 26, 2013 #9
    I understand the difference between quantum and classical. But even in quantum, we can define the spin direction by asking whether it is the eigenstate of a spin operator in that direction. If the state is an eigenstate of the spin operator in some direction, then I say the spin is pointing in that direction. I don't mean to suggest that here "pointing in that direction" has the same meaning as in the classical world. So a spin (1,0) is "pointing in +z" direction in my language and (cosθ/2, sinθ/2) is pointing in (sinθ, 0, cosθ) direction.

    Now, the point of the analogy to the classical case is only to demonstrate the idea that the 10 fundamental operators have a certain commutation relations mathematically speaking, so the same operation, boosting a spin in a perpendicular direction should have the same result, namely I think a spin (1,0) should still be (1,0), if it changes, we can always find out "the direction" as I defined above. Again, the direction here just means it's an eigenstate of a spin operator in that direction.
     
  11. Jun 26, 2013 #10
    What's the Pauli-Lyubarsky vector?
     
  12. Jun 26, 2013 #11

    Avodyne

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    In general, under a boost, the quantum state of a particle with spin undergoes a "Wigner rotation". See section 17 of http://physics.uoregon.edu/~soper/qft/wigner.pdf [Broken] The final result is eq.(111). Note that there is an implicit sum over s' in this equation (and in many earlier equations). This means that, in general, the boosted state does not have a definite z-component of the spin.
     
    Last edited by a moderator: May 6, 2017
  13. Jun 26, 2013 #12
    Notice that in eq.(111) the operation is on a state that's already moving with momentum k, which can be seen as boosted from k0, when you do another boost from k to [itex]\Lambda[/itex]k, if they are not collinear, then the result will differ from a boost from k0 directly to [itex]\Lambda[/itex]k by a rotation. In other words, the Wigner rotation appears when you have at least two non-collinear boosts, but this is not what I am after. In my case, it is a single boost from rest frame.

    I am asking if the spin direction would change if I boost it from rest frame in a direction that's not parallel to the original spin direction. If I were to boost a spinor that's already moving with momentum k to some other frame moving with k', then I would have a Wigner rotation. But here, it is simply a question of whether "one single boost" from rest frame changes the spin orientation. At this moment, I think the answer looks like no, according the long reasoning I posted above.
     
    Last edited by a moderator: May 6, 2017
  14. Jun 26, 2013 #13

    Avodyne

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    In that case, the non-rotation is true by definition; we define the spinor us(p) to be the boost of us(0), as in eq.(107), after which the author says "By definition, this boost operation on this state does not mix states with different spin indices."

    But this doesn't automatically tell us what happens in a Stern-Gerlach apparatus; we would need to figure out how the boosted spin interacts with the inhomogeneous magnetic field. Offhand I don't know of an analysis of Stern-Gerlach with relativistic particles.

    EDIT: found one, see post #16 below.
     
    Last edited: Jun 26, 2013
  15. Jun 26, 2013 #14

    dextercioby

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    As a side note, it would be nice if prof. Soper didn't mistake O(3) for SO(3) in his (useful) notes referenced above.
     
  16. Jun 26, 2013 #15
    Thanks for pointing that out! I think that makes sense.

    I was thinking about the "contradiction" I brought up in my previous post:

    I think I figured out what's wrong with it. I was assuming that the operation only applies to the spin but not the motion of the particle. But it seems that the spinor is an object that not only reflects the spin orientation but also the momentum with which the particle is moving. So the operation should also apply to the motion. In other words, the above mentioned two sets of operations don't give the same final state, as illustrated in the attached picture, so there's no contradiction with the commutation relations. The black arrow represents the momentum, and blue arrow represents the spin direction. This also says that although the final spin directions are the same, but because the particle is in a different reference frame (moving with different momentum [itex]\mathbf{P}'[/itex] rather than [itex]\mathbf{P}[/itex]), the spinor states are not the same. In other words, for each different momentum, there's a different set of spin basis, so that even for spins that have the same orientation, if they are in different inertia frames, their spinor states will be different.
     

    Attached Files:

  17. Jun 26, 2013 #16

    Avodyne

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    It seems that Stern-Gerlach for relativistic particles is rather involved. Here's a recent paper that seems to have done it carefully: http://arxiv.org/abs/1208.6434
     
  18. Jun 28, 2013 #17

    DrDu

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